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kari74 [83]
3 years ago
13

If a cart of 2 kg mass has a force of 8 newtons exerted on it, what is its acceleration?

Physics
1 answer:
maria [59]3 years ago
3 0
This is an example of the Newton`s Second Law:
F = m * a
a = F / m
F = 8 N, m = 2 kg.
a = 8 N : 2 kg
Answer:
a = 4 m/s²
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the higher the ramp the less distance it will travel

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The process of combining two small nuclei into one nucleus of larger mass is called _____.
fiasKO [112]
The process of splitting one large nucleus into
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Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, stra
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Answer:

how large a magnetic field would you experience = 8.16 x 10∧-4T

Explanation:

I = 20KA = 20,000A

r = 4.9 m

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4 0
3 years ago
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A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute
Amiraneli [1.4K]

it can be said that  the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

  • what is the speed of the east wind?.

<h3> the speed of the east wind</h3>

Generally the equation for the distance  is mathematically given as

BA=3000sin60

BA=2598.07m

Therefore

the speed of the east wind

V_w=\frac{BA}{120*60}\\\\V_w=\frac{2598.07}{120*60}

v=0.3608

For more information on this visit

brainly.com/question/22568180

7 0
2 years ago
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